Challenge 9: Baby Names

Author

Kyle Bistrain

show the code

library(tidyverse)
library(broom)
library(knitr)
library(DT)
library(kableExtra)
library(tinytex)
library(moderndive)

show the code

baby_names_data <- read_csv(here::here("supporting_artifacts",
                               "extended learning",
                               "StateNames_A.csv"
                               )
                    )

baby_names_data |> 
  datatable()

Make a summary table of the number of babies named “Allison” for each state and the sex of the baby. Specifically, each state should be its own row and each sex should have its own column. Additionally, if there were no babies born with the name “Allison” for that combination of year, state, & sex the table should have a value of 0 (not an NA).

show the code

baby_names <- baby_names_data |> 
  rename(sex_at_birth = `Gender`) |> 
  filter(`Name` == 'Allison') 
  
  baby_names |> 
  group_by(sex_at_birth,State) |> 
  summarise(sum = sum(`Count`), .groups = "drop") |> 
  pivot_wider(names_from = `sex_at_birth`,
              values_from = `sum`
              ) |> 
  mutate(M = coalesce(M, 0)) |> 
  kable(format = "html",
        col.names = c("State", 
                      "Frequency of Babies Sex_at_Birth: F", 
                      "Frequency of Babies Sex_at_Birth: M"
                      ),
        align = "ccc",
        booktabs = TRUE
        ) |> 
    kable_styling(font_size = 12)

State	Frequency of Babies Sex_at_Birth: F	Frequency of Babies Sex_at_Birth: M
AK	232	0
AL	1535	0
AR	1198	0
AZ	1880	0
CA	12413	0
CO	1594	0
CT	1099	0
DC	321	0
DE	294	0
FL	4455	0
GA	3257	0
HI	183	0
IA	1477	0
ID	451	0
IL	5110	0
IN	3067	0
KS	1283	0
KY	1905	20
LA	1209	0
MA	2218	0
MD	2229	0
ME	340	0
MI	4014	0
MN	2374	0
MO	2882	0
MS	817	0
MT	226	0
NC	3435	0
ND	285	0
NE	807	0
NH	412	0
NJ	3052	0
NM	399	0
NV	729	0
NY	5747	0
OH	5487	0
OK	1421	0
OR	1186	0
PA	4307	0
RI	306	0
SC	1228	0
SD	376	0
TN	2488	0
TX	10192	0
UT	1125	0
VA	3220	0
VT	135	0
WA	1956	0
WI	2367	0
WV	813	0
WY	142	0

source: https://sparkbyexamples.com/r-programming/replace-na-values-with-zero-in-r-dataframe/

You should have seen in the table above that “Allison” is a name given overwhelmingly to babies assigned female at birth. So, filter the data to include only babies assigned Female at birth.
Make a visualization showing how the popularity of the name “Allison” has changed over the years. To be clear, each year should have one observation–the total number of Allisons born that year.

show the code

 baby_names <- baby_names|> 
  filter(`sex_at_birth` != 'M') |> 
  group_by(Year) |> 
  summarise(year_count = sum(`Count`), .groups = "drop")
 baby_names |> 
  ggplot(mapping = aes(x = Year, y = year_count)) +
  geom_bar(stat = "identity") +
  geom_text(aes(label = year_count), color = "blue", vjust=0) +
  ggtitle("Count of Babies named Allison \n with female sex_at_birth") + 
  labs(x = "Year", y = "")

4 Modeling the Number of Allisons

Create a linear model with the year as the explanatory variable, and the number of Allisons as the response. Similar to #4, each year should have one observation–the total number of Allisons born that year.
show the code
```
baby_names_model <- baby_names |> 
  lm(year_count ~ Year, data = _)
```

Visualize the regression model.

show the code

baby_names |> 
  ggplot(aes(x = Year, y = year_count)) +
  geom_point() +
  geom_smooth(formula = y ~ x, method = "lm", se = FALSE) +
  labs(x = "Year", 
       y = "") +
  ggtitle('year_count ')

Write out the estimated regression equation.

show the code

get_regression_table(baby_names_model) |> 
  kable(format = "html")

term	estimate	std_error	statistic	p_value	lower_ci	upper_ci
intercept	209689.761	42971.505	4.880	0	118594.240	300785.282
Year	-101.519	21.427	-4.738	0	-146.942	-56.096

\[ \hat{y} = 209689.761- (101.519*Year) \]

\[ \hat{y} = \] predicted number of babies birthed in a year with the name Allison and sex_at_birth Female

Plot the residuals of the model, that is, the actual values minus the predicted values. Comment on the residuals - do you see any patterns?

show the code

baby_names_model |> 
  augment() |> 
  ggplot(aes(x = Year, y = .resid)) +
  geom_point() +
  ggtitle('Residuals') +
  labs(x = "Year",
       y = "")

From the graph of the Residuals, I does appears that there is a U type pattern in the graph. This means that the residuals are likely not normally distributed. This does not satisfy the condition and therefore doesn’t satisfy all the linear model conditions.

What do you conclude from this model? Is my name not cool anymore?

Therefore, from this model we do not satisfy the normally distributed residuals condition therefore we can’t make a conclusion based on our linear model. From early analysis it may appear that the name Allison is becoming less popular over time, we can’t make any claims.

5 Spelling by State

In middle school I was so upset with my parents for not naming me “Allyson”. Past my pre-teen rebellion, I’m happy with my name and am impressed when baristas spell it “Allison” instead of “Alison”. But I don’t have it as bad as my good friend Allan!

Narrow the A name dataset (downloaded previously) down to only male-assigned babies named “Allan”, “Alan”, or “Allen”. Make a plot comparing the popularity of these names over time.

show the code

baby_names_p5 <- baby_names_data |> 
  rename(sex_at_birth = `Gender`) |> 
  filter(`Name` %in% c('Allan','Alan','Allen'),
         `sex_at_birth` != 'F'
         )


  baby_names_p5 |> 
    group_by(Year,Name) |> 
    summarise(year_count = sum(`Count`), .groups = "drop") |> 
    ggplot(aes(x = Year, y = year_count, color = Name)) + 
    geom_point() +
    ggtitle('year_count') +
    labs(x = "Year", y = "")

In California, Allan’s spelling of his name is the least common of the three

but perhaps it’s not such an unusual name for his home state of Pennsylvania. Compute the total number of babies born with each spelling of “Allan” in 2000, in Pennsylvania and in California. Specifically, each spelling should be its own column and each state should have its own row. Similar to before, a 0 (not an NA) should be used to represent locations where there were no instances of these names.

show the code

baby_names_p5 |> 
  filter(`State` %in% c("CA","PA"),
         `Year` == 2000
         ) |> 
  group_by(State,Name) |> 
  summarise(year_count = sum(`Count`), .groups = "drop") |> 
  pivot_wider(names_from = `Name`,
              values_from = `year_count`
              ) |> 
   kable(format = "html",
         col.names = c("State", 
                      "Frequency of Alan",
                      "Frequency of Allan",
                      "Frequency of Allen"),
         align = "ccc",
         booktabs = TRUE
         ) |> 
    kable_styling(font_size = 12)

State	Frequency of Alan	Frequency of Allan	Frequency of Allen
CA	579	131	176
PA	51	12	56

Convert your total counts to overall percents. That is, what was the percent breakdown between the three spellings in CA? What about in PA?

show the code

baby_names_p5 |> 
  filter(`State` %in% c("CA","PA"),
         `Year` == 2000
         ) |> 
  group_by(State,Name) |> 
  summarise(sum_cnt = sum(Count), .groups = "drop") |> 
  pivot_wider(names_from = `Name`,
              values_from = `sum_cnt`
              ) |> 
  mutate(`row_sum` = `Alan` + `Allan` + `Allen`,
         `Alan` = (`Alan`/`row_sum`)*100,
         `Allan` = (`Allan`/`row_sum`)*100,
         `Allen` = (`Allen`/`row_sum`)*100
         ) |> 
  select(-5) |> 
  kable(format = "html",
        col.names = c("State", 
                      "percent named Alan",
                      "percent named Allan",
                      "percent named Allen"),
        align = "ccc",
        booktabs = TRUE
        ) |> 
    kable_styling(font_size = 12)

State	percent named Alan	percent named Allan	percent named Allen
CA	65.34989	14.78555	19.86456
PA	42.85714	10.08403	47.05882

After producing the table, the births of babies with the name Allan is still the most uncommon in both California and Pennsylvania.